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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote Probability Question This topic has 2 expert replies and 1 member reply Probability Question Timer 00:00 Your Answer A B C D E Global Stats Difficult Hi I need help with the following question. Four people each will roll a fair die once. What is the probability that at least two will roll the same number? A) 50% B) 65% C) 80% D) 72% E) 40% The problem can easily be solved by making all the people roll the same number then subtracting the result from 1. However, I wanted to understand the problem by adding up all the cases i.e two roll the same number or three roll the same number etc. but i am not getting the correct answer. Can any expert help me find the mistake? Case 1 Two people throw the same number and the other two each roll a different number. e.g 4456 6/6 * 1/6 *5/6 * 4/6 *4!/3!2! I am not sure how to calculate all the permutations of this. To arrange four numbers i multiplied by 4! then divided by 3! to eliminate double counting and again by 2! to remove the same digit cases i.e 44 55. Case 2 two roll the same number and the other two roll the same but different from the one that was rolled by the first two people. e.g 3344 6/6* 1/6* 5/6* 1/6 *4/2! 2! 2! Here I arrange the four numbers then divided by 2! three times to remove double digits already arranged in the product and to eliminate two same digit groups. again i am not sure if this is correct way to count all permutations of this case. Case 3 three roll the same number and the third rolls a different number eg 4445 6/6 * 1/6*1/6*5/6 * 4!/ 2!*3! Case 4 All roll the same number this is easy 6/6^4 Adding up all the cases does not yield the right answer. Please help me find the mistake. Thanks Jac GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15344 messages Followed by: 1864 members Upvotes: 13060 GMAT Score: 790 Top Reply Jacwis wrote: Hi I need help with the following question. Four people each will roll a fair die once. What is the probability that at least two will roll the same number? A) 50% B) 65% C) 80% D) 72% E) 40% I wanted to understand the problem by adding up all the cases All possible rolls: Since there are 6 number options for each roll, we get: 6*6*6*6 = 1296 Case 1: Exactly 2 rolls are the same (such as 1123) From 4 rolls, the number of ways to choose a pair to yield the same number = 4C2 = (4*3)/(2*1) = 6. Number options for this pair = 6. (Any of the 6 numbers on the die) Number options for the third roll = 5. (Any of the 5 remaining numbers on the die) Number options for the last roll = 4. (Any of the 4 remaining numbers on the die). To combine these options, we multiply: 6*6*5*4 = 720 Case 2: Two pairs (such as 1122) Number of ways to divide 4 rolls into 2 pairs = 3. (First way: 1st and 2nd rolls are the same; 3rd and 4th rolls are the same but different from the 1st and 2nd.) (Second way: 1st and 3rd rolls are the same; 2nd and 4th rolls are the same but different from the 1st and 3rd.) (Third way: 1st and 4th rolls are the same; 2nd and 3rd rolls are the same but different from the 1st and 4th.) Number options for first pair = 6. (Any of the 6 numbers on the die) Number options for the second pair = 5. (Any of the 5 remaining numbers on the die) To combine these options, we multiply: 3*6*5 = 90 Case 3: Exactly 3 numbers are the same From 4 rolls, the number of ways to choose three to yield the same number = 4C3 = (4*3*2)/(3*2*1) = 4. Number options for these 3 rolls = 6. (Any of the 6 numbers on the die) Number options for the fourth roll = 5. (Any of the 5 remaining numbers on the die) To combine these options, we multiply: 4*6*5 = 120 Case 4: All 4 numbers are the same Number options = 6. (Any of the 6 numbers on the die) Case 1 + Case 2 + Case 3 + Case 4 = 720 + 90 + 120 + 6 = 936 Thus: P(at least 2 people roll the same number) = 936/1296 â‰ˆ 0.72 = 72% The correct answer is D. MUCH easier to solve as follows: P(at least 2 numbers are the same) = 1 - P(all 4 numbers are different) The first person can roll any of the 6 numbers. P(2nd number is different from the 1st) = 5/6 P(3rd number is different from the first 2 numbers) = 4/6 P(4th number is different from the first 3 numbers) = 3/6 To combine these probabilities, we multiply: 5/6 * 4/6 * 3/6 = 5/6 * 2/3 * 1/2 = 5/18 Thus: P(at least 2 numbers are the same) = 1 - 5/18 = 13/18 â‰ˆ 0.72 = 72% _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. Newbie | Next Rank: 10 Posts Joined 21 Apr 2019 Posted: 2 messages Thanks Mitch for the solution I have some confusion, in the second case in your post, 1122 . When we select two positions for a pair from the four positions, it can be done in 4C2=6 ways. However, when we can place any pair in 6 ways then the second pair can be placed in the remaining two positions in only 1 way. Isn't it so?Why are there 3 ways to roll the two pairs? For example 1122 can be arranged 4!/2!2! which is 6. All the cases of 1122 are 1122 1212 1221 2211 2121 2112 What is wrong with this reasoning. Please, elaborate. Thanks GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15344 messages Followed by: 1864 members Upvotes: 13060 GMAT Score: 790 Jacwis wrote: two roll the same number and the other two roll the same but different from the one that was rolled by the first two people. e.g 3344 6/6* 1/6* 5/6* 1/6 *4/2! 2! 2! Here I arrange the four numbers then divided by 2! three times to remove double digits already arranged in the product and to eliminate two same digit groups. again i am not sure if this is correct way to count all permutations of this case. When we select two positions for a pair from the four positions, it can be done in 4C2=6 ways. However, when we can place any pair in 6 ways then the second pair can be placed in the remaining two positions in only 1 way. Isn't it so?Why are there 3 ways to roll the two pairs? For example 1122 can be arranged 4!/2!2! which is 6. All the cases of 1122 are 1122 1212 1221 2211 2121 2112 What is wrong with this reasoning The approach above counts the following: Cases that include 11: 1122, 1133, 1144, 1155, 1166 Cases that include 22: 2211, 2233, 2244, 2255, 2266 Cases that include 33: 3311, 3322, 3344, 3355, 3366 Cases that include 44: 4411, 4422, 4433, 4455, 4466 Cases that include 55: 5511, 5522, 5533, 5544, 5566 Cases that include 66: 6611, 6622, 6633, 6644, 6655 Notice the following: The cases in red have all been double-counted. 1122 will yield the same outcomes as 2211. 1133 will yield the same outcomes as 3311. 1144 will yield the same outcomes as 4411. 1155 will yield the same outcomes as 5511. 1166 will yield the same outcomes as 6611. By extension, EVERY case above has been double-counted. To account for the double-counting, we must divide by 2: (6/6* 1/6* 5/6* 1/6)(4!/2!2!)/2 _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! 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